Answer:
Area: 135 ft^2
Perimeter: 50 ft
Step-by-step explanation:
area:
take the rectanle so length 12 x 9 = 108 so that is the length of the rectangle and now we need to find that of the triangle left over
subract 18 - 12 = 6 so that is the base of the trianle and we know the side length is 9 so plus it in A = (9)(6)/2
A = 54/2
A = 27
add 27 + 108 to get the total area
135
perimeter:
18 + 9 + 11 + 12 = 50
Supplementary angles
x+20+5x+10=180
6x+30=180
6x=150
x=25
Adding or subtracting a constant from each value of a distribution changes the mean, but not the standard deviation.
The standard deviation will equal 3.
The largest possible volume of the given box is; 96.28 ft³
<h3>How to maximize volume of a box?</h3>
Let b be the length and the width of the base (length and width are the same since the base is square).
Let h be the height of the box.
The surface area of the box is;
S = b² + 4bh
We are given S = 100 ft². Thus;
b² + 4bh = 100
h = (100 - b²)/4b
Volume of the box in terms of b will be;
V(b) = b²h = b² * (100 - b²)/4b
V(b) = 25b - b³/4
The volume is maximum when dV/db = 0. Thus;
dV/db = 25 - 3b²/4
25 - 3b²/4 = 0
√(100/3) = b
b = 5.77 ft
Thus;
h = (100 - (√(100/3)²)/4(5.77)
h = 2.8885 ft
Thus;
Largest volume = [√(100/3)]² * 2.8885
Largest Volume = 96.28 ft³
Read more about Maximizing Volume at; brainly.com/question/1869299
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Answer:
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Step-by-step explanation:
